Angles — Paper 4 · IGCSE Mathematics - International 0607

7 A ship sails 65 km on a bearing of 310° from A to B. It then changes course and sails 40 km on a bearing of 250° from B to C. The ship then returns to A. (a) On the diagram, sketch the path of the ship from A. On your diagram show the bearings and distances. North A [3] (b) Find angle ABC. … [1] (c) Calculate AC and show that it rounds to 91.8 km, correct to the nearest tenth of a kilometre. [3] (d) Find the bearing of C from A. … [4]

11 marks

Mark scheme: 7(a) Correct skettch showing bearingsb 3 B1B for 310° bearingb apprrox correct (2270 to 360) aand and distancees markedm B1B for 250° bearingb apprrox correct (1180 to 270) aand markedm B1B for distannces correctlyy marked 7(b) 120 1 7(c) 402 + 652 – 22×40×65×coos their 120 M1 theirt 120 muust be betweeen 0 and 180 40 2 + 655 2 − [ ]2 AllowA cos1220 = 2 × 400 × 65 91.78 to 91.779 A2 A1A for 8425 or 5 337 7(d) 288 or 287.88... 4 40sin ( their120 ) M2M for oe 91.8 sinθ sin(thheir 120) oro M1 for = oe 404 991.8 IfI cosine rulee used, M2 fofor explicit exxpression or M1M for impliicit. A1A for 22.2 oro 22.16 to 222.17… IfI 0 scored SC2 for answwer 108 or 1007.8…

This question in 0607/42 May/June 2017

2 (a) D A C 68° NOT TO E SCALE B In the diagram, ABC is a triangle and AB is parallel to DE. Angle BCA = 68˚and DE = DC. (i) Find angle BAC. Angle BAC = … [2] (ii) scalene equilateral isosceles right-angled Choose one word from the list to complete the statement. Triangle ABC is … [1] (b) Calculate the interior angle of a regular 20 sided polygon. … [3] (c) C B P NOT TO SCALE A Q R In the diagram, angle A = angle P and angle B = angle Q. (i) Explain why angle C = angle R. … [1] (ii) AB = 8 cm, AC = 5 cm, BC = 9 cm and PR = 3 cm. (a) Complete the statement. Triangle ABC is … to triangle PQR [1] (b) Calculate QR. QR = … cm [2]

10 marks

Mark scheme: 2(a)(i) 44 2 M1 for [angle BAC or DEC =] 180 – 2 × 68, soi by angle CDE = 44 or M1 for angle BAC = their angle CDE 2(a)(ii) isosceles 1 2(b) 162 3 360 180 × (20 − 2) M2 for 180 – or 20 20 360 or M1 for or 180 × (20 – 2) 20 2(c)(i) Angle sum of triangle oe 1 2(c)(ii)(a) similar 1 2(c)(ii)(b) 5.4 2 5 9 M1 for = oe 3 QR

This question in 0607/43 May/June 2017

8 North B NOT TO SCALE 17 km 142° C North 4 km A Rani sails in a boat race around a triangular course. She sails from A to B to C and then directly back to A. B is due north of C. (a) Find the bearing Rani sails on from C to A. … [1] (b) Show that AB = 20.3 km, correct to 1 decimal place. [3] (c) Calculate the bearing of B from A. … [3] (d) Rani starts the race at 08 57 and returns to A at 12 33. Calculate the average speed of her boat in km/h. … km/h [3]

10 marks

Mark scheme: 8(a) 218 1 8(b) 42 + 172 – 2 × 4 × 17 × cos142 M2 M1 for implicit cosine rule 20.30… A1 8(c) 007 or 006.92 to 006.98 3 4sin142 M2 for sin B = oe 20.3 4 20.3 or M1 for = oe sin B sin142 OR 17sin142 M2 for sin A= oe 20.3 17 20.3 or M1 for = oe sin A sin142 8(d) 11.5 or 11.47… 3 B1 for 3 h 36 min or 3.6 h seen 4 + 17 + 20.3 M1 for their 3.6

This question in 0607/42 Feb/March 2021

7 In this question all lengths are in centimetres. (a) C B 8x° ( x + 5 )° NOT TO SCALE A In triangle ABC, AC = BC, angle ABC = ( x + 5)° and angle ACB = 8x° . Find the value of x. x = … [3] (b) NOT TO ( p - 2) SCALE ( p + 1) The diagram shows a rectangle with sides of length ( p + 1) and ( p - 2) . The area of the rectangle is 90 cm2 . Find the value of p. p = … [4] (c) ( y - 1) ( y - 4) NOT TO SCALE 30° The diagram shows a right-angled triangle. Find the value of y. y = … [3] (d) 13 ( w + 1 ) NOT TO SCALE ( 2w + 3) The diagram shows a right-angled triangle with sides of length ( w + 1) , ( 2w + 3) and 13. Work out the area of the triangle. … cm2 [6]

16 marks

Mark scheme: 7(a) 17 3 M2 for x + 5 + 8 x + x + 5 = 180 oe or M1 for angle A = x + 5 7(b) 10.1 or 10.10... 4 B3 for correct sketch indicating roots −−( 1) ± ( − 1) 2 − 4(1)( − 92) or for oe 2(1) or B2 for p 2 − 2 p + p − 2 [ = 90] or better or M1 for ( p + 1)( p − 2) [ = 90] 7(c) 7 3 M2 for 2(y – 4) = y – 1 or better y − 4 or M1 for = sin30 y − 1 If 0 scored SC1 for sin 30 = 0.5 7(d) 2.04 oe 6 B4 for (5 w − 1)( w + 3) or correct sketch indicating roots − 14 ± 14 2 − 4(5)( −3) or 2(5) or B3 for 5 w 2 + 14 w − 3 = 0 and M1 for correct calculation of area of triangle with their positive w OR 2 2 M1 for ( w + 1) + ( 2 w + 3 ) = 13 B1 for w 2 + w + w + 1 oe or 4 w 2 + 6 w + 6 w + 9 oe and M1 for correct calculation of area of triangle with their positive w

This question in 0607/43 May/June 2021

8 D 7 cm NOT TO C SCALE 18 cm A 13 cm 16 cm B (a) Calculate angle BCA and show that it rounds to 59.57°, correct to 2 decimal places. [3] (b) Find the area of quadrilateral ABCD. … cm2 [3] (c) Find the shortest distance from A to BC. … cm [2] (d) D is due north of B. Find the bearing of B from C. … [6]

14 marks

Mark scheme: 8(a) 18 2 + 132 − 16 2 M2 M1 for 16 2 = 182 + 132 −×2 18 × 13cos(...) 2 × 18 × 13 59.574 to 59.575 A1 8(b) 164 or 163.8 to 163.9 3 1 M1 for × 18 × 7 oe 2 1 M1 for × 18 × 13 × sin59.57 oe 2 8(c) 15.5 or 15.52... 2 distance M1 for sin 59.57 = oe 18 8(d) 191 or 190.5 to 190.6 6 M2 for 7 2 + 132 −×2 7 × 13cos(90 + 59.57) or B1 for [angle BCD =] 149.57 7sin(90 + 59.57) M2 for theirBD theirBD 7 or M1 for = sin(90 + 59.57) sin DBC M1 for 180 + their DBC oe

This question in 0607/43 May/June 2021

4 (a) a° 65° b° NOT TO 30° SCALE c° The diagram shows two straight lines crossing two parallel lines. Find the values of a, b and c. a = … b = … c = … [3] (b) L D y° 20°20° 70° E w° NOT TO SCALE K 30° u° C v° x° A z° B A, B, C, D and E are points on the circle. KL is a tangent to the circle at E. AC = AD. Find the values of u, v, w, x, y and z. u = … x = … v = … y = … w = … z = … [6]

9 marks

Mark scheme: 4(a) [ a = ] 65 3 B1 for each [ b = ] 85 [ c = ] 95 4(b) [u = ] 70 6 B1 for each [v = ] 30 [w = ] 80 FT 180 – their u – their v [x = ] 20 [y = ] 50 FT 150 – their x – their w [ z = ] 60 FT 110 – their y

This question in 0607/42 Feb/March 2022

8 North NOT TO B 120° SCALE North 65 km C 55° A The diagram shows the route of a ship between three ports, A, B and C. The bearing of B from A is 055° and the bearing of C from B is 120°. BC = 65 km . The ship takes 7 hours to sail from A to B. It sails at a speed of 20 km/h. (a) Find the distance AB. … km [1] (b) Show that angle ABC = 115° . [1] (c) (i) Calculate the distance CA. … km [3] (ii) Calculate the bearing of A from C. … [4] (d) The ship takes 3.6 hours to sail from B to C. It then sails from C to A at a speed of 21.5 km/h. Find the average speed for the complete journey from A to B to C and back to A. … km/h [3]

12 marks

Mark scheme: 8(a) 140 1 8(b) 360 – (120 + 125) or 60 + 55 1 or 180 + 55 – 120 8(c)(i) 178 or 177.5... 3 M2 for ((their140) 2  65 2  2  ( their140)  65  cos115) OR M1 for (their 140)2 + 652 – 2 × (their 140) × 65 × cos115 8(c)(ii) 254 or 255 or 254.3 to 254.5... 4 their140sin115 M2 for sin[C ]  oe their178 sin[ C ] sin115 or M1 for  oe their140 their178 A1 for 45.18 to 45.63 M1 for 360 – 60 – their C oe calculated as answer 8(d) 20.3 or 20.26 to 20.31... 3 their140  65  their178 M2 for their178 7  3.6  21.5 their178 or M1 for 21.5 totaldistance or for clear indication of totaltime

This question in 0607/43 May/June 2022

9 C D NOT TO 118° SCALE 5 m 12 m 35° A B 16 m (a) B is due east of A. Find the bearing of A from C. … [2] (b) Calculate the area of triangle ABC. … m2 [2] (c) Calculate angle CAD. Angle CAD = … [4] (d) Calculate the length of the straight line BD. … m [3]

11 marks

Mark scheme: 9(a) 235 2 M1 for 180  55 or for 360 – 125 or for 270 – 35 or for 35 or 55 or 125 or 145 correctly indicated at C. 9(b) 55.1 or 55.06... 2 1 M1 for  12  16 sin 35 oe 2 9(c) 40.4 or 40.41... 4 5sin118 M2 for [sinC = ] 12 12 5 or M1 for  oe sin118 sinC M1 dep for 180 – 118 – their C dependent on sine rule used to find angle. 9(d) 15.5 or 15.51... nfww 3 M2 for 5 2  16 2  2 5 16cos(35 theirA) or M1 for 52 + 162 – 2 516cos(35 + theirA) A1 for 241 or 240.6 to 240.7...

This question in 0607/41 May/June 2023

5 (a) The diagram shows a regular pentagon with sides of 10 cm and centre O. B 10 cm A C NOT TO O SCALE E D (i) Find angle AOB. Angle AOB = … [1] (ii) Show that OA = 8. 51 cm correct to 3 significant figures. [3] (iii) Find the area of the pentagon. … cm2 [2] (b) V NOT TO 18 cm SCALE B A C O E 10 cm D The regular pentagon in part (a) is the base of a pyramid. The sloping edges, VA, VB, VC, VD, and VE, are each of length 18 cm. (i) Calculate the perpendicular height, VO, of the pyramid. VO = … cm [3] (ii) Calculate the volume of the pyramid. … cm3 [2] (iii) A geometrically similar pyramid has volume 1500 cm 3. Calculate the length of a side of the base of this pyramid. … cm [3]

14 marks

Mark scheme: 5(a)(i) 72 1 5(a)(ii) 5 M2  1 5 oe oe M1 for sin  their 72   1   2  OD sin  their 72   2  8.506 to 8.507 A1 5(a)(iii) 172 or 172.0 to 172.2 2 1 M1 for × 8.512 × sin(their 72) oe 2 1 or  8.51  (10or5)sin54 oe 2 1 or  (10or5)  5tan54 oe 2 5(b)(i) 15.9 or 15.86... 3 M2 for 182 – 8.512 or M1 for VO2 + 8.512 = 182 5(b)(ii) 909 to 913 2 1 M1 for  (their172)  (their15.9) 3 5(b)(iii) 11.8 or 11.79 to 11.82 3 1500 M2 for 10 × 3 oe their(b)(ii) 1500 their(b)(ii) or M1 for 3 or 3 their(b)(ii) 1500 their (b)(ii)  10  3 or   oe 1500  x 

This question in 0607/42 May/June 2023

5 B North NOT TO SCALE 123 m 154 m North A 27° 183 m C 106° D The diagram shows a field ABCD, with a straight path AC. The bearing of C from A is 122° . (a) Calculate the bearing of D from C. … [3] (b) Show that angle ABC = 81.9° correct to one decimal place. [3] (c) Find the total area of the field ABCD. … m2 [5]

11 marks

Mark scheme: 5(a) 255 cao 3 B2 for 75 correctly referenced at C or D OR B1 for angle ACD = 47 B1 for angle ACN(orth) = 58 or ACS(outh) = 122 5(b) 1232 + 154 2 − 1832 M2 M1 for 1832 = 1232 + 1542 – 2 × 123 × [cos] = 154 × cos[…] 2  123  154 81.87... A1 5(c) 15 200 or 15 150 to 15 161 5 183sin27 M2 for CD = sin106 CD 183 or M1 for = sin27 sin106 OR 183sin their 47 M2 for AD = sin106 AD 183 or M1 for = sin their 47 sin106 AND M1 for 1 × 154 × 123 × sin 81.9 2 M1 for 1 × 183 × their CD × sin their47 2 or 1 × 183 × their AD × sin 27 2 or 12 × their CD × their AD × sin 106

This question in 0607/42 Feb/March 2024

7 (a) F 30° E NOT TO SCALE G 130° A B C D ABCD is a straight line and EC and BF meet at G. BE is parallel to CF and GF = CF . Angle ABE = 130° and angle BFC = 30° . Find (i) angle FCD Angle FCD = … [2] (ii) angle FBC Angle FBC = … [1] (iii) angle BGE. Angle BGE = … [2] (b) B NOT TO SCALE C X A D A, B, C and D are points on the circle. AC and BD meet at X. (i) Show that triangles AXB and DXC are similar. Give a reason for each statement you make. … … … … … [2] (ii) AX = 5 cm, XC = 2 cm and XD = 4 cm. Find the length of BD. BD = … cm [3]

10 marks

Mark scheme: 7(a)(i) 50 2 B1 for angle BCF = 130 or angle EBC = 50 soi by angle EBG = 30 and angle GBC = 20 or angle FCG = 75 and angle GCB = 55 7(a)(ii) 20 1 7(a)(iii) 75 2 180 − 30 M1 for 2 7(b)(i) 2 from 2 B1 for 2 pairs correct with no/incorrect Angle AXB = Angle DXC reasons and conclusion [Vertically] opposite angles or for one pair correct with reason. Angle ABX = Angle DCX Angles in same segment. Angle BAX = Angle CDX Angles in same segment. And conclusion AA[A] 7(b)(ii) 6.5 3 B2 for BX = 2.5 ... 5 or M1 for = oe 2 4

This question in 0607/42 Feb/March 2024